Suppose that $M$ is a completely reducible module over a ring $R$, that is, $$M=\bigoplus_{\alpha\in J}M_\alpha$$ where each $M_\alpha$ is an irreducible $R$-module (irreducible means it has no proper non-trivial submodule).
Prove that $M$ is Artinian if and only if it is Noetherian.
This is Problem 7 (on page 122) of $\S$3.5 in Jacobson's Basic Algebra II (2nd edition). Jacobson earlier proves the following characterisation of completely reducible modules (on page 121), but I am not able to see how to use it.
Theorem 3.10. The following conditions in a module $M \neq 0$ are equivalent:
$M = \sum M_\alpha$, where the $M_\alpha$ are irreducible.
$M$ is completely reducible.
The lattice $L(M)$ of submodules of $M$ is complemented, that is, for every submodule $N$ of $M$ there exists a submodule $N'$ of $M$ such that $M = N \oplus N'$.
Any hints on how to solve this? Some intuition also?
If $M$ is either Artinian or Noetherian,then the index set $J$ is finite.every irreducible module is Noetherian and Artinian.